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3 years ago

(6x – 10y + 12) – 3

Mathematics

1 answer:

Margarita [4]3 years ago

8 0

6x-10y+9

Step-by-step explanation:

(6x – 10y + 12) – 3

6x-10y+12-3

subtract 12-3

6x-10y+9

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What is the 4th term of the geometric sequence when the first term is 6 and the common ratio is 7?

RUDIKE [14]

Answer:

2,058

Step-by-step explanation:

first term : a_1 = 6

common ratio r = 7

a_n = (a_1) * r ^(n-1)

a_4 = 6 * 7^(4-1)

a_4 = 6 *7^3

a_4 = 6 * 343 = 2,058

5 0

3 years ago

(HELP ASAP)

wariber [46]

Answer:

45 km^2 (C)

Step-by-step explanation:

Isolate the smaller square -

We only need two side lengths, take 3km and 3km.

3×3 = 9km^2, that's the area of the smaller square.

Isolate the larger square -

Take 6km and the other 6km

6×6 = 36km^2

Add the square's areas together to get the whole area -

36 + 9 = 45km^2

All you need to do to solve these problems is to isolate the shapes and find their area's, and finally add them together.

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3 years ago

Suppose that an individual has a body fat percentage of 19.7% and weighs 157 pounds. How many pounds of his weight is made up of

mixas84 [53]

Answer:

Step-by-step explanation:

6 0

3 years ago

Find the volume of the solid formed by revolving the region bounded by LaTeX: y = \sqrt{x} y = x and the lines LaTeX: y = 1 y =

Strike441 [17]

Answer:

The volume is:

$\displaystyle\frac{37\pi}{10}$

Step-by-step explanation:

See the sketch of the region in the attached graph.

We set the integral using washer method:

$\displaystyle\int_a^b\pi r^2dx$

Notice here the radius of the washer is the difference of the given curves:

$x-\sqrt{x}$

So the integral becomes:

$\displaystyle\int_1^4\pi(x-\sqrt{x})^2dx$

We solve it:

Factor $\pi$ out and distribute the exponent (you can use FOIL):

$\displaystyle\pi\int_1^4x^2-2x\sqrt{x}+x\,dx$

Notice: $x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}$

So the integral becomes:

$\displaystyle\pi\int_1^4x^2-2x^{3/2}+x\,dx$

Then using the basic rule to evaluate the integral:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{2x^{5/2}}{5/2}+\frac{x^2}{2}\right|_1^4$

Simplifying a bit:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{4x^{5/2}}{5}+\frac{x^2}{2}\right|_1^4$

Then plugging the limits of the integral:

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(4)^{5/2}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Taking the root (rational exponents):

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(2)^{5}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Then doing those arithmetic computations we get:

$\displaystyle\frac{37\pi}{10}$

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3 years ago

A bat and a ball cost one dollar and ten cents in total. The bat costs a dollar more than the ball. How much does the ball cost?

lozanna [386]

10 cents becuase you just - it

5 0

3 years ago

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